Gravitational potential energy (roller coaster)

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  • #1
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Homework Statement


A 1000kg roller coaster, with its passengers, starts from rest at point A( h= 9.5 m) on a friction track whose profile is shown in the diagram above at right.
a) what is its maximum speed?
b) with what speed does the roller coaster arrive at point e?
c) What constant braking force would have to be applied to the roller coaster at point E, to bring it to rest in a horizontal distance of 5.0 m
Given data (from diagram)
Point A- 9.5m
Point B- 6.5m
Point C- 9.2m
Point D- 0.50m
Point E- 5.5m

Homework Equations


Ep= mgh



The Attempt at a Solution


I used the gravitational potential energy equation to find gravitational potential energy at point A. I don't know if I need to get it for all the points! Also I don't understand where to start in this equation. I also wrote down the givens.
 

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Answers and Replies

  • #2
NascentOxygen
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Did you attach the diagram? Oh, I see, you provided the relevant heights.

So it will be going fastest at the lowest point, assuming that is D. Is there another equation you know, relating energy to a moving body?
 
  • #3
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For part A, Ep = mgh.
Conservation of energy law states that the Ep at the top is equal to the Ek at the bottom. Ek = .5mv^2
So, mgh = .5mv^2, where v is the velocity at the bottom.

For the next part, I would use the conservation of energy law again. Except this time, you have velocity at the point where potential energy is the highest, so you have to account for the total mechanical energy of the system. At point B, which i'm assuming is at the top of another hill at 6.5m, Ep + Ek = Ep + Ek. The right side of the equation is for point B, and the left side of the equation is for the bottom of the hill after point A. mgh + .5mv^2 = mgh + .5mv^2., solve for v, and do so until you get to point e.

For C, F=ma
you know the mass, so now you need acceleration
V = SQRT(2AD)
solve for A, since D is 5 and V is the velocity you're going at point e that you just found

then, F = ma
 
  • #4
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Kinetic energy, work. I also know that Kinetic energy is equal to work.?
 
  • #5
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The height in part a are the same right?
 
  • #6
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V = SQRT(2AD). What does sqrt stand for? Sorry I have never used that!
 
  • #7
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For part A, Ep = mgh.
Conservation of energy law states that the Ep at the top is equal to the Ek at the bottom. Ek = .5mv^2
So, mgh = .5mv^2, where v is the velocity at the bottom.

For the next part, I would use the conservation of energy law again. Except this time, you have velocity at the point where potential energy is the highest, so you have to account for the total mechanical energy of the system. At point B, which i'm assuming is at the top of another hill at 6.5m, Ep + Ek = Ep + Ek. The right side of the equation is for point B, and the left side of the equation is for the bottom of the hill after point A. mgh + .5mv^2 = mgh + .5mv^2., solve for v, and do so until you get to point e.

For C, F=ma
you know the mass, so now you need acceleration
V = SQRT(2AD)
solve for A, since D is 5 and V is the velocity you're going at point e that you just found

then, F = ma
a) is it 99.05 m/s (also do i use the same height)
b) for this I have to do every height there is?
c) how is D 5? in the diagram you only have 5.5 and/or 0.5!
 
  • #8
NascentOxygen
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How did you arrive at 99.05? Without explanation, who is to say whether you are right or not??

In part c, the track levels out at a height of 5.5 metres. You want to bring the car to a halt in 5 metres along this level section, being the end of track.
 
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  • #9
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uh I found 99.05 by solving mgh= hv^2.
(1000) (9.81) (0.5) = 0.5v^2
4905/0.5= 0.5v^2/0.5
Square(9810)= square(v^2)
99.05m/s= v
 
  • #10
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okay, now that I looked at the diagram lets approach this in another way.
Let's find the Ep and Ek and velocities at each point, that way we can calculate the correct max speed, which will probably be at the lowest height, and get the right values for each point.

so for part A, there is no velocity, so it's all potential.
Ep = mgh = 93100 Joules

Point B, the total mechanical energy of point A must equal the total mechanical energy of point B, B has Ep and Ek, because there is height and velocity

mgh = .5mv^2 + mgh
(1000)(9.8)(9.5) = (.5)(1000)(v^2) + (1000)(9.8)(6.5)
93100 = 500v^2 + 63700
29400 = (.5)(1000)v^2
58.8 = v^2
v = 7.668m/s at point B

Moving onto point C, Energy at point B = .5mv^2 + mgh at point C
(.5)(1000)(7.668^2) + (1000)(9.8)(6.5) = .5(1000)v^2 + (1000)(9.8)(9.2)
v at point c = 15.065 m/s

same thing for point D,
.5mv^2 + mgh = .5mv^2 + mgh
v = 19.936m/s

and finally for point E,
.5mv^2 + mgh = .5mv^2 + mgh
v = 17.305m/s

A) so, our max speed is at point D, which is 19.936
B) we just got is, 17.305
C) now we're moving in the horizontal. Our initial speed is 17.305 and our final speed is 0.
F=MA
to find acceleration, or in this case,V = SQRT(2AD)
17.305 = the square root of (2*5*A)
so, A = 29.947
F = 29947N

am I correct?
 
  • #11
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sorry, F = MA , = 518.232N
 
  • #12
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How did you get 518.232N? Like I know F= MA but what acceleration did you use?
 
  • #13
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... A = 29.947? the one we found on the last step
 

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